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Let $ f\left( z \right) = \sin z $ and $ g\left( z \right) = \cos z $ . If $ * $ denotes a composition of the function then show that $ \left( {f + ig} \right) * \left( {f - ig} \right) = i{e^{ - {e^{iz}}}} $ .
 $ i = \sqrt { - 1} $

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Hint: Since we have to show $ \left( {f + ig} \right) * \left( {f - ig} \right) = i{e^{ - {e^{iz}}}} $ and for this we will write $ \left( {f + ig} \right) * \left( {f - ig} \right) $ in the composition form and it will be $ F\left( {G\left( z \right)} \right) $ . From here $ F\left( z \right) = f + ig $ and $ G\left( z \right) = f - ig $ . Then we will solve both the composite equation and then in the end putting the values we will get the solution.
Formula used:
The algebraic formula is given by,
 $ \cos z - i\sin z = {e^{ - iz}} $
 $ i = \sqrt { - 1} $
Here, $ i $ will be the iota.

Complete step-by-step answer:
Here, in this question, we have the values of the function given as $ f\left( z \right) = \sin z $ and $ g\left( z \right) = \cos z $ .
And we have to prove $ \left( {f + ig} \right) * \left( {f - ig} \right) = i{e^{ - {e^{iz}}}} $
Now taking the LHS of the upper function, we get
 $ \Rightarrow \left( {f + ig} \right) * \left( {f - ig} \right) $
And it can also be written in the composition form, so it will be
 $ \Rightarrow F\left( z \right) * G\left( z \right) $
And it will be equal to
 $ \Rightarrow F\left( {G\left( z \right)} \right) $
Since, $ F\left( z \right) = f + ig $
So, on substituting the values, we get
 $ \Rightarrow F\left( z \right) = \left( {\sin z + i\cos z} \right) $
Multiplying and dividing the above equation with the same function, $ i $
We get,
 $ \Rightarrow F\left( z \right) = \left( {\sin z + i\cos z} \right)\dfrac{i}{i} $
And on solving it, we get
 $ \Rightarrow F\left( z \right) = \dfrac{{i\sin z + {i^2}\cos z}}{i} $
Since we know $ i = \sqrt { - 1} $ , hence
 $ \Rightarrow F\left( z \right) = \dfrac{{i\sin z - \cos z}}{i} $
Taking the negative sign common, we get
 $ \Rightarrow F\left( z \right) = - \dfrac{1}{i}\left( {\cos z - i\sin z} \right) $
Also, from the formula, we know that $ \cos z - i\sin z = {e^{ - iz}} $ . So by using it we get
 $ \Rightarrow F\left( z \right) = - \dfrac{1}{i}{e^{ - iz}} $
Again multiplying and dividing the above equation with the same function, $ i $
We get
 $ \Rightarrow F\left( z \right) = - \dfrac{1}{i} \cdot \dfrac{i}{i}{e^{ - iz}} $
And on solving it we get
 $ \Rightarrow F\left( z \right) = - \dfrac{{1 \cdot i}}{{{i^2}}}{e^{ - iz}} $
Hence, it will be equal to
 $ \Rightarrow F\left( z \right) = i{e^{ - iz}} $
Now we will solve, $ G\left( z \right) = f - ig $
So, on substituting the values, we get
 $ \Rightarrow G\left( z \right) = \left( {\sin z - i\cos z} \right) $
Multiplying and dividing the above equation with the same function, $ i $
We get,
 $ \Rightarrow G\left( z \right) = \left( {\sin z - i\cos z} \right)\dfrac{i}{i} $
And on solving it, we get
 $ \Rightarrow G\left( z \right) = \dfrac{{i\sin z - {i^2}\cos z}}{i} $
Since we know $ i = \sqrt { - 1} $ , hence
 $ \Rightarrow G\left( z \right) = \dfrac{{i\sin z + \cos z}}{i} $
Also, from the formula, we know that $ \cos z - i\sin z = {e^{ - iz}} $ . So by using it we get
\[ \Rightarrow G\left( z \right) = \dfrac{{{e^{iz}}}}{i}\]
Again multiplying and dividing the above equation with the same function, $ i $
We get
 $ \Rightarrow G\left( z \right) = - \dfrac{1}{i} \cdot \dfrac{i}{i}{e^{iz}} $
And on solving it we get
\[ \Rightarrow G\left( z \right) = - \dfrac{{1 \cdot i}}{{ - 1}}{e^{iz}}\]
Hence, it will be equal to
 $ \Rightarrow G\left( z \right) = - i{e^{iz}} $
So from the upper part of the solution,
 $ \Rightarrow F\left( {G\left( z \right)} \right) = i{e^{i \cdot \left( {G\left( z \right)} \right)}} $
And on substituting the values, we get
 $ \Rightarrow F\left( {G\left( z \right)} \right) = i{e^{i \cdot \left( { - i{e^{iz}}} \right)}} $
By using identities, we get the equation as
 $ \Rightarrow F\left( {G\left( z \right)} \right) = i{e^{ - {i^2}{e^{iz}}}} $
And it can also be written as
 $ \Rightarrow F\left( {G\left( z \right)} \right) = i{e^{{e^{iz}}}} $
Since, we can see that LHS is equal to RHS,
Therefore, it is proved that $ \left( {f + ig} \right) * \left( {f - ig} \right) = i{e^{ - {e^{iz}}}} $ will be equal.
So, the correct answer is “ $ \left( {f + ig} \right) * \left( {f - ig} \right) = i{e^{ - {e^{iz}}}} $ ”.

Note: Solving composite functions means we have to find the composition of two functions. By rewriting the composite function in a different form we will be able to reach close to the answer. And lastly, by substituting the values we get it.